Why take $\sqrt {A^ \dagger A}$ in the polar decomposition of matrix A?

Question

In the Polar Decomposition section in Nielsen and Chuang (page 78 in the 2002 edition), there is a claim that any matrix $A$ will have a decomposition $UJ$ where $J$ is positive and is equal to $\sqrt{A^\dagger A}$.

To me, it looks like the reason the $\sqrt{A^\dagger A}$ has been chosen as the positive matrix is because its null space is the same as A, and on the other vectors, it's action is the same as A upto a unitary, which is the unitary we multiply be to finally get $A = UJ$.

But what is the need to take $\sqrt{A^\dagger A}$ ? Won't $A^\dagger A$ do the same job?

Answer 1

The point is that $U$ is supposed to be unitary, which means it preserves lengths. For any vector $v$, $\|\sqrt{A^\dagger A} v\| = \|A v\|$. That would not be true of $A^\dagger A v$. For one thing, it scales wrong.

Answer 2

Polar decomposition of matrices should generalise polar decomposition of (complex) numbers. For a complex number $A$, $A^\dagger A = |A|^2$ so you need the square root to get the proper decomposition $A = |A| U$ which really is a polar decomposition in an elementary geometric way. Indeed, in this case, $U = cos(\theta) + i \sin(\theta)$ where $\theta$ is an argument of $A$.